what are the values of m when $m(3^{2m}+3) \equiv 0 \mod 28$?

Question

Can you help me with this problem. I want to know what are the values of m which makes $m(3^{2m}+3) \equiv 0 \mod 28$.

Answer 1

Clearly, one solution is $$28|m$$

Else we need $$3^{2m}+3\equiv0\pmod{28}\iff 3^{2m-1}\equiv-1\pmod{28}\text{ as }(3,28)=1$$

Now as $28=4\cdot7,$ and $3\equiv-1\pmod4\implies3^{2m-1}\equiv-1\pmod4$

So the problem reduces to finding $m$ such that $3^{2m-1}\equiv-1\pmod7$

Now, $3^2\equiv2\pmod7,3^3\equiv-1$

So $2m-1$ needs to be any odd multiple of $3$ i..e, $2m-1=3(2r+1)$ where $r$ is any integer

$\implies 2m=6r+4\iff m=3r+2$